Solve for $x$ : $ 6|x - 2| + 2 = -4|x - 2| + 3 $
Explanation: Add $ {4|x - 2|} $ to both sides: $ \begin{eqnarray} 6|x - 2| + 2 &=& -4|x - 2| + 3 \\ \\ { + 4|x - 2|} && { + 4|x - 2|} \\ \\ 10|x - 2| + 2 &=& 3 \end{eqnarray} $ Subtract ${2}$ from both sides: $ \begin{eqnarray} 10|x - 2| + 2 &=& 3 \\ \\ { - 2} &=& { - 2} \\ \\ 10|x - 2| &=& 1 \end{eqnarray} $ Divide both sides by ${10}$ $ \dfrac{10|x - 2|} {{10}} = \dfrac{1} {{10}} $ Simplify: $ |x - 2| = \dfrac{1}{10}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x - 2 = -\dfrac{1}{10} $ or $ x - 2 = \dfrac{1}{10} $ Solve for the solution where $x - 2$ is negative: $ x - 2 = -\dfrac{1}{10} $ Add ${2}$ to both sides: $ \begin{eqnarray} x - 2 &=& -\dfrac{1}{10} \\ \\ {+ 2} && {+ 2} \\ \\ x &=& -\dfrac{1}{10} + 2 \end{eqnarray} $ Change the ${ + 2}$ to an equivalent fraction with a denominator of $10$ $ x = - \dfrac{1}{10} {+ \dfrac{20}{10}} $ $ x = \dfrac{19}{10} $ Then calculate the solution where $x - 2$ is positive: $ x - 2 = \dfrac{1}{10} $ Add ${2}$ to both sides: $ \begin{eqnarray} x - 2 &=& \dfrac{1}{10} \\ \\ {+ 2} && {+ 2} \\ \\ x &=& \dfrac{1}{10} + 2 \end{eqnarray} $ Change the ${ + 2}$ to an equivalent fraction with a denominator of $10$ $ x = \dfrac{1}{10} {+ \dfrac{20}{10}} $ $ x = \dfrac{21}{10} $ Thus, the correct answer is $x = \dfrac{19}{10} $ or $x = \dfrac{21}{10} $.